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I have the derivatives of two functions, $\frac{df(t)}{dt}$ and $\frac{dg(t)}{dt}$. I would like to calculate the derivative $\frac{df(t)}{dg(t)}$.

This is a reasonably simple problem:

\begin{equation} \frac{df(t)}{dg(t)} = \frac{\frac{df(t)}{dt}}{\frac{dg(t)}{dt}} \end{equation}

However, suppose I only have access to a variable $x$ which defines $t$:

$x = \cosh(t-0.5)$, $x\in\{0,1\}$

The inverse equation requires two branches to describe properly:

$t = \pm\cosh^{-1}(x)+0.5$, $t\in\{-0.5,0.5\}$

So now I would like to calculate:

\begin{equation} \frac{df(t(x))}{dg(t(x))} = \frac{\frac{df(t(x))}{dt(x)}}{\frac{dg(t(x))}{dt(x)}} \end{equation}

I will surely have two solutions, for which there will be degenerate solutions at each value of $x$: one solution for each domain in $t$.

I have never dealt with such a situation before.

Is my understanding correct, and if so, what is the correct way of rendering $\frac{df(t(x))}{dg(t(x))}$? Is what I am attempting even a valid thing to do?

Jack Rolph
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  • What is with the absolute value? $$\frac{df(t)}{dg(t)} = \dfrac{\frac{df(t)}{dt}}{\frac{dg(t)}{dt}}$$ when the two right-side derivatives exist and the derivative of $g(t)$ is not $0$. – Paul Sinclair Nov 26 '21 at 04:11

1 Answers1

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Assuming you know the derivatives of $f(t),g(t)$, the expression you wish to find is a function of $t$ too. $$F(t)=\frac{\,df(t)}{\,dg(t)}=\frac{f'(t)}{g'(t)}$$ The only use of $x$ is to find the value of $t$ at which you wish to find $F(t)$. And of course,for $x=\cosh(t-0.5)$, you get two valid values of $t$, hence a valid branch must be specified.

flash1_maths
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